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Let $A$ be the localisation of $R = \mathbb{Z}[X,Y]/(XY-9)$ at $\mathfrak{m} = (x,y,3)$. I wondered if this ring is maybe Noetherian, Artin, or regular.

Noetherian: Notice that $\mathbb{Z}$ is a PID, and thus Noetherian. By Hilberts' basis theorem we find that $\mathbb{Z}[X,Y]$ is also Noetherian. This implies that $R$ is Noetherian. Since $A = S^{-1}R$ with $S = R-\mathfrak{m}$ we find that $A$ is also Noetherian.

Artin: Notice that $A$ is Artin if and only if $A$ is Noetherian and of dimension $0$. Now notice that $A$ is a local ring with one maximal ideal, and consequently we have the sequence of prime ideals $0\subset\mathfrak{m}_{A}$ of length $1$. So the dimension of $A$ is at least $1$. Thus $A$ can not be Artin.

Regular: With showing that $A$ is regular or not I struggle. Since I think that one has to calculate the dimension of $A$, and then see whether $\mathfrak{m}_{A}$ can be generated by $d$ elements where $d$ is the dimension of $A$.

JanBakfiets1
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1 Answers1

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We have $R = \mathbb{Z}[X,Y]/(XY-9)=\mathbb{Z}[x,y],\;\; \mathfrak m=\langle x,y,3\rangle, \;\;A=R_\mathfrak m$.
Let us denote by $M={\frak m} A_{\frak m}$ the maximal ideal of $A$, so that the residue field of $A$ is $\frac A M=\frac {R} {\mathfrak m}=\mathbb F_3$.
The ring $R$ has dimension $2$ by Krull's Hauptidealsatz and thus $A=R_\mathfrak m$ has dimension $2$ too, since we have the chain $\langle 0\rangle \subsetneq \langle x,3 \rangle \subsetneq \langle x,y,3 \rangle=\mathfrak m$ of length $2$ of prime ideals in $R$ giving rise to a chain of length $2$ of prime ideals in $A=R_{\mathfrak m}$.
Now $A=R_\mathfrak m$ is not regular because $\operatorname {dim}_\mathbb {F_3} (\frac{M}{M^2})=\operatorname {dim}_\mathbb {F_3} (\frac{\mathfrak m}{\mathfrak m^2})=3\gt 2$ and this implies that $R$ is not regular, since the definition of a regular ring is that all its localizations are regular.

Georges Elencwajg
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